A parallel plate capacitor of capacitance 2 F is charged to a potential V . energy stored in capacit

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2. Electric Potential and Capacitance

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A parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E _2$. The ratio $E _2 / E _1$ is

A

$2: 1$

B

$1: 2$

C

$1: 4$

D

$2: 3$

(JEE MAIN-2023)

Solution

Initially

$Q _1= CV =(2) V$

$E _1=1 / 2 CV ^2=1 / 2(2) V ^2= V ^2$

Finally

Charge on each capacitor, $Q _2=\frac{ Q _1}{2}=\frac{2 V }{2}= V$

$E _2=2\left(\frac{1}{2} \frac{ Q _2^2}{ C }\right)=\frac{ V ^2}{2} \quad \therefore \frac{ E _2}{ E _1}=\frac{1}{2}$

Standard 12

Physics

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